Q. 4.45 A student is getting ready to ta... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 4: Q. 4.45 (page 166)

A student is getting ready to take an important oral examination and is concerned about the possibility of having an 鈥渙n鈥� day or an 鈥渙ff鈥� day. He figures that if he has an on the day, then each of his examiners will pass him, independently of one another, with probability $8$ , whereas if he has an off day, this probability will be reduced to $4$ . Suppose that the student will pass the examination if a majority of the examiners pass him. If the student believes that he is twice as likely to have an off day as he is to have an on the day, should he request an examination with $3$ examiners or with $5$ examiners?

Short Answer

Expert verified

The probability of passing is more, if $3$ examiners, So he should prefer $3$ examiners.

Step by step solution

Given Information

A student is getting ready to take an important oral examination and is concerned about the possibility of having an 鈥渙n鈥� day or an 鈥渙ff鈥� day.

Probability of off day $= \frac{2}{3}$

Probability of on day $= \frac{1}{3} .$

Solution of the Problem

If he request $3$ Examiners then:

$P [pass if 2 or more emaminers pass him]$

$= \frac{2}{3} [(\begin{array}{l} 3 \\ 2 \end{array}) (0.4)^{2} 路 (0.6) + (\begin{array}{l} 3 \\ 3 \end{array}) (0.4)^{3}] + \frac{1}{3} [(\begin{array}{l} 3 \\ 2 \end{array}) (0.8)^{2} 路 (0.2) + (\begin{array}{l} 3 \\ 3 \end{array}) (0.8)^{3}]$

$= \frac{2}{3} [0.288 + 0.064] + \frac{1}{3} [0.384 + 0.512]$

We get,

$= 0.2347 + 0.2987$

$= 0.533 .$

Computation of the Probability

If he request $5$ Examiners then:

$P [pass if 3 or more examiners pass him]$

$P (pass) = \frac{2}{3} [(\begin{array}{l} 5 \\ 3 \end{array}) (0.4)^{3} 路 (0.6)^{2} + (\begin{array}{l} 5 \\ 4 \end{array}) (0.4)^{4} 路 (0.6) + (\begin{array}{l} 5 \\ 5 \end{array}) (0.4)^{5}]$

$+ \frac{1}{3} [(\begin{array}{l} 5 \\ 3 \end{array}) (0.8)^{3} 路 (0.2)^{2} + (\begin{array}{l} 5 \\ 4 \end{array}) (0.8)^{4} 路 (0.2) + (\begin{array}{l} 5 \\ 5 \end{array}) (0.8)^{5}]$

We get,

$= \frac{2}{3} [0.2304 + 0.0768 + 0.01024] + \frac{1}{3} [0.2048 + 0.4096 + 0.32768]$

$= 0.2117 + 0.314$

$= 0.526$

Final Answer

The probability of passing is more, if $3$ examiners, So he should prefer $3$ examiners.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Given Information

Solution of the Problem

Computation of the Probability

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Calculus

Logic and Functions

Geometry

Mechanics Maths

Study anywhere. Anytime. Across all devices.